Lösungsvorschlag

Mantelfläche eines Rotationskörpers (siehe Buch S.257-258):

$M = 2 π \int_{a}^{b} f (x) \sqrt{1 + f^{'} (x)^{2}} d x$

Funktion ist linear mit Anstieg $\frac{r}{h}$

$f (x) = \frac{r}{h} x, f^{'} (x) = \frac{r}{h}$

$M = 2 π \int_{0}^{h} \frac{r}{h} x \sqrt{1 + {(\frac{r}{h})}^{2}} d x$

$= 2 π \frac{r}{h} \sqrt{1 + {(\frac{r}{h})}^{2}} \int_{0}^{h} x d x$

$= 2 π \frac{r}{h} \sqrt{1 + {(\frac{r}{h})}^{2}} h^{2} \frac{1}{2}$

$= π \frac{r}{h} \sqrt{1 + {(\frac{r}{h})}^{2}} h^{2}$

$= π \frac{r}{h} \sqrt{\frac{h^{2}}{h^{2}} + \frac{r^{2}}{h^{2}}} h^{2}$

$= π \frac{r}{h} \frac{\sqrt{r^{2} + h^{2}}}{h} h^{2}$

$= π r \sqrt{r^{2} + h^{2}}$

$= π r s$ (weil $s = \sqrt{r^{2} + h^{2}}$ )

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